Integration operator appears inside an integration operator as I try $~\mathcal{L}\left[x \cdot \sin^{}\left(x\right) \right]\left(s\right)~$ $$  A:=\mathcal{L}\left[x \cdot \sin^{}\left(x\right) \right]\left(s\right) \tag{1}  $$
$$=\lim_{\beta\to\infty}\int_{0}^{\beta}\left(x\cdot\sin^{}\left(x\right)\right)\cdot\exp\left(-sx\right)\,dx$$
$$ = \lim_{\beta\to\infty}\int_{0}^{\beta}\left(x\right)\cdot \underbrace{\left( \sin^{}\left(x\right) \exp\left(-sx\right) \right)}_{\text{This part is to be integrated} }   \,dx  $$
$$ = \lim_{ \beta \to \infty} \left\{ \left[ x \cdot \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right)  \,dx   \right)  \right]_{0}^{\beta} - \int_{0 }^{\beta } \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right)  \,dx  \right)  \,dx   \right\}  $$
$$  \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right)  \,dx = - \frac{ s \cdot  \exp\left(-sx\right)   }{ \left( s^2+1 \right)    } \left( \sin^{}\left(x\right) + \frac{1}{s}\cos^{}\left(x\right)  \right) +\text{const} \tag{2}   $$
Should I have written the derivation of the above equation?
$$ A= \lim_{ \beta \to \infty} \left\{ \left[ -\frac{  x  \cdot s \cdot \exp\left(-sx\right)   }{ \left( s^2+1 \right)    } \left( \sin^{}\left(x\right) + \frac{1}{s}\cos^{}\left(x\right)  \right)  \right]_{0}^{\beta} -  \int_{0 }^{\beta } \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right)  \,dx  \right)  \,dx    \right\}  $$
$$ = \lim_{ \beta \to \infty} \left\{ \underbrace{\left[ -\frac{  x  \cdot s   }{ \left( s^2+1 \right) e^{sx}   } \left( \sin^{}\left(x\right) + \frac{1}{s}\cos^{}\left(x\right)  \right)  \right]_{0}^{\beta}}_{\text{About}~\beta~ \text{,it converges to }~0   }  -  \int_{0 }^{\beta } \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right)  \,dx  \right)  \,dx    \right\}  $$
$$ = -\lim_{ \beta \to \infty} \underbrace{\int_{0 }^{\beta } \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right)  \,dx   \right)  \,dx}_\text{What can be done at here?}    $$
 A: Just proceed calculations using equations of laplace transformations of $~ \sin^{}\left(x_{}\right) ~~,~~ \cos^{}\left(x\right)  ~$
$$  A=\mathcal{L}\left[x \cdot \sin^{}\left(x\right) \right]\left(s\right) \tag{1}  $$
$$ = -\lim_{ \beta \to \infty} \int_{0 }^{\beta } \left( \int_{ }^{ } \sin^{}\left(x\right) \exp\left(-sx\right)  \,dx   \right)  \,dx    $$
$$ = -\lim_{ \beta \to \infty} \int_{0 }^{\beta } \left(- \frac{  s  }{ s^2+1   }  \right) \frac{  1  }{ e^{sx}   } \left( \sin^{}\left(x\right) + \frac{1}{ s } \cos^{}\left(x\right)   \right)    \,dx    $$
$$ = \frac{  s  }{ s^2+1   } \lim_{ \beta \to \infty} \int_{ 0}^{ \beta} \frac{1}{ e^{sx} } \left( \sin^{}\left(x\right) + \frac{1}{ s } \cos^{}\left(x\right)  \right)  \,dx     $$
$$ = \frac{  s  }{ s^2+1   } \lim_{ \beta \to \infty} \int_{ 0}^{\beta }  \left( \sin^{}\left(x\right) \exp\left(-sx\right)  + \frac{1}{ s } \cos^{}\left(x\right) \exp\left(-sx\right)   \right)  \,dx     $$
$$ = \frac{  s  }{  s^2+1  } \left( \mathcal{L}\left[\sin^{}\left(x\right) \right]\left(s\right)  + \frac{1}{ s }  \mathcal{L}\left[\cos^{}\left(x\right) \right]\left(s\right)\right)   $$
$$ = \frac{  s  }{ s^2+ 1   }  \left( \frac{  1  }{  s^2+1  } + \frac{1}{ s }  \frac{  s  }{ s^2+1   }   \right)  $$
$$ = \frac{  s  }{ s^2+ 1   } \left( \frac{  2  }{  s^{2}+1  }  \right)   $$
$$ = \frac{  2s  }{ \left( s^{2}+1 \right) ^2   }  $$
